Suppose that the order of a (mod p) is odd and equals to 2k+1 for some positive integer k. This means that a^(2k+1) ≡ 1 (mod p). Since p is prime and a is relatively prime to p, by Fermat’s Little Theorem, a^(p-1) ≡ 1 (mod p). Therefore, we have

a^(2k+1) ≡ a^(p-1) (mod p)

Subtracting 1 from both sides, we get

a^(2k+1) – 1 ≡ a^(p-1) – 1 (mod p)

Using the difference of squares factorization, we can rewrite the left side as

(a^k + 1)(a^k – 1) ≡ 0 (mod p)

Since a is relatively prime to p, both a^k + 1 and a^k – 1 are relatively prime to p. Therefore, they must be either 0 or 1 (mod p). However, they cannot both be 0 (mod p) because that would mean that a^k ≡ 1 (mod p), which would contradict the assumption that the order of a (mod p) is odd. Therefore, one of them must be 1 (mod p) and the other must be 0 (mod p). Without loss of generality, suppose that a^k + 1 ≡ 0 (mod p). Then we have

a^k ≡ -1 (mod p)

Squaring both sides, we get

a^(2k) ≡ 1 (mod p)

Thus,

(a/p) = (a^(2k) * (1/p)) = 1 * (1/p) = 1

Therefore, we have shown that if the order of a (mod p) is odd, then (a/p) = 1.